You can find all the previous posts about Vedic Mathematics below:

Introduction to Vedic Mathematics

A Spectacular Illustration of Vedic Mathematics

10's Complements

Multiplication Part 1

Multiplication Part 2

Multiplication Part 3

Multiplication Part 4

Multiplication Part 5

Multiplication Special Case 1

Multiplication Special Case 2

Multiplication Special Case 3

Vertically And Crosswise I

Vertically And Crosswise II

Squaring, Cubing, Etc.

Subtraction

Division By The Nikhilam Method I

Division By The Nikhilam Method II

Division By The Nikhilam Method III

Division By The Paravartya Method

Digital Roots

Straight Division I

Straight Division II

Vinculums

Divisibility Rules

Simple Osculation

Multiplex Osculation

Solving Equations 1

Solving Equations 2

Solving Equations 3

Solving Equations 4

Mergers 1

Mergers 2

Mergers 3

To set the stage, consider an equation of the type below:

3/(x + 2) + 2/(x + 1) + 4/(x + 7) = 9/(x + 4)

We see that there are not two, but three terms on the left hand side of the equation, but there is only one term on the right hand side of the equation. Moreover, the sum of numerators on the left hand side (3 + 2 + 4) is equal to the numerator on the right hand side (9). Also, the coefficients of the unknown in all the terms is unity.

Now, look at the series of manipulations below:

3/(x + 2) + 2/(x + 1) + 4/(x + 7) = 9/(x + 4) becomes

3/(x + 2) - 3/(x + 4) + 2/(x + 1) - 2/(x + 4) + 4/(x + 7) - 4/(x + 4) = 0 becomes

3*(x + 4 - x - 2)/[(x + 2)*(x + 4) + 2*(x + 4 - x - 1)/[(x + 1)*(x + 4) + 4*(x + 4 - x - 7)/[(x + 7)*(x + 4)] = 0 becomes

[1/(x + 4)] * [3*2/(x + 2) + 2*3/(x +1) -4*3/(x + 7)] = 0 becomes

6/(x + 2) + 6/(x + 1) - 12/(x + 7) = 0 becomes

6/(x + 2) + 6/(x + 1) = 12/(x + 7)

We now see that the right hand side term has been merged with the left hand side, and we are left with an equation that is once again capable of undergoing another merger to get the solution (the numerators on the left hand side, 6 and 6, add up to the numerator on the right hand side, 12, and the coefficients of the unknown in all three terms are unity again).

Performing the merger operation on the resulting equation gives us a final equation of:

30/(x + 2) + 36/(x + 1) = 0

This then gives us a solution of x = -17/11. One can verify that this indeed is the correct solution to both the original given equation as well as the intermediate equations derived after the first and second mergers.

Since the solution procedure for the given equation involved performing two mergers one after the other, this procedure is referred to as a multiple merger. In fact, under the right conditions, it is possible to string together any number of mergers to solve an equation. The question then becomes: what are the right conditions?

To answer that question, let us consider a general equation as below:

p/(x + a) + q/(x + b) + r/(x + c) = s/(x + d)

Let us now assume that p + q + r = s. Thus we can perform a merger operation to begin with, and eliminate s and d, giving us:

p*(d - a)/(x + a) + q*(d - b)/(x + b) + r*(d - c)/(x + c) = 0

To be able to perform a second merger and solve the equation, the condition that has to be satisfied is as below:

Either p*(a - d) = q*(d - b) + r*(d - c) or

q*(b - d) = p*(d - a) + r*(d - c) or

r*(c - d) = p*(d - a) + q*(d - b)

If any of the above conditions are satisfied, then a second merger can be performed and the equation can be solved using merger. Assuming the third condition is satisfied, we can rewrite the equation after the first merger as below:

p*(d - a)/(x + a) + q*(d - b)/(x + b) = r*(c - d)/(x + c)

Because the third condition is true, we know that the sum of the numerators on the left hand side of the above equation is equal to the numerator on the right hand side of the equation. Therefore, we can perform another merger operation to eliminate r*(c - d) and c. This will give us:

p*(d - a)*(c - a)/(x + a) + q*(d - b)*(c - b)/(x + b) = 0

This can then be solved using the formula we derived in the first lesson on mergers here.

The extension of this methodology to more than 2 mergers must now be obvious to the reader. At each step, after the performance of a merger, the remaining terms must satisfy the conditions for another merger to be performed. The conditions imply that the sum of all but one of the terms must be equal to the negative of one of the terms. Depending on which term is thus isolated, a merger would once again be performed on the equation until the equation is reduced to two terms, at which point it can be solved using the formula we derived earlier.

Checking for these conditions to be satisfied when the number of terms increases becomes more and more difficult, and it may not be possible to do this mentally without a whole lot of practice. This lesson is therefore more of an exploration of the possibilities rather than a concrete series of mental steps to solve equations using this method.

However, it may be worth going through the first few steps of a merger operation just to see if the equation can be solved with much less effort than the process of cross-multiplying, expanding out all the terms, etc. If the merger works out, even if it is not fully successful all the way down to getting the equation down to two terms, any terms that are eliminated are terms you don't have to consider when solving the equation manually using the process of cross-multiplication, expansion and collection of terms, etc. And a few eliminated terms also reduces the degree of the resulting equation when you expand the equation out fully, and that also reduces the effort involved in solving it.

If you are even more adventurous, note that this method works when the coefficients of the unknown are all the same, but not unity in each of the terms of the given equation. The final equation we get would then be solved using the formula we derived in this earlier lesson.

Moreover, the equation may also require some preliminary transformation, as we have seen in the previous lesson, before a merger can be attempted. Unfortunately, these will almost surely make the equation too complicated to attempt to solve mentally, so I am not going to delve into these in detail. The mechanics of these operations have been explained in detail in previous lessons, so it is possible to extend and apply the tests and transformations we derived earlier on paper to simplify the equation before solving it mentally as a final step. Good luck, and happy computing!

And I wish you all a very happy and prosperous new year too!!

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